A car traveling initially at +5.7 m/s acceler-
ates uniformly at the rate of +0.79 m/s2 for a
distance of 262 m.
What is its velocity at the end of the accel-
eration?
Answer in units of m/s
What is its velocity after it accelerates for 132
m?
Answer in units of m/s
What is its velocity after it accelerates for 65
m?
Answer in units of m/s
To find the final velocity of the car at the end of acceleration, we can use the equation:
v^2 = u^2 + 2aS
where:
v = final velocity
u = initial velocity
a = acceleration
S = distance
For the first question, the initial velocity (u) is +5.7 m/s, the acceleration (a) is +0.79 m/s^2, and the distance (S) is 262 m. Plugging in these values into the equation, we have:
v^2 = (5.7 m/s)^2 + 2*(0.79 m/s^2)*(262 m)
Simplifying this equation, we get:
v^2 = 32.49 m^2/s^2 + 413.44 m^2/s^2
v^2 = 445.93 m^2/s^2
To find v, we take the square root of both sides:
v = √445.93 m^2/s^2
v ≈ 21.12 m/s
Therefore, the velocity at the end of acceleration is approximately 21.12 m/s.
For the second question, since the distance for the accelerated motion is given as 132 m, we can repeat the calculation using the same values of initial velocity (u) and acceleration (a), but replacing the distance (S) with 132 m:
v^2 = (5.7 m/s)^2 + 2*(0.79 m/s^2)*(132 m)
Simplifying this equation, we get:
v^2 = 32.49 m^2/s^2 + 209.76 m^2/s^2
v^2 = 242.25 m^2/s^2
Taking the square root of both sides, we find:
v = √242.25 m^2/s^2
v ≈ 15.57 m/s
Therefore, the velocity after accelerating for 132 m is approximately 15.57 m/s.
For the third question, using the same initial velocity (u) and acceleration (a), but replacing the distance (S) with 65 m:
v^2 = (5.7 m/s)^2 + 2*(0.79 m/s^2)*(65 m)
Simplifying this equation, we get:
v^2 = 32.49 m^2/s^2 + 102.94 m^2/s^2
v^2 = 135.43 m^2/s^2
Taking the square root of both sides, we find:
v = √135.43 m^2/s^2
v ≈ 11.63 m/s
Therefore, the velocity after accelerating for 65 m is approximately 11.63 m/s.
To find the velocity at the end of the acceleration, you can use the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken. In this case, the initial velocity (u) is +5.7 m/s, the acceleration (a) is +0.79 m/s^2, and the distance (s) is given as 262 m.
To find the time taken (t), you can use the equation:
s = ut + (1/2)at^2
Rearranging the equation, you get:
2as = u^2 + 2at
Substituting the known values, you can solve for t:
2(0.79)(262) = (5.7)^2 + 2(0.79)t
524.36 = 32.49 + 1.58t
491.87 = 1.58t
t = 311.16 s
Now that we have the time taken, we can substitute it back into the first equation to find the velocity at the end of the acceleration:
v = u + at
v = 5.7 + 0.79(311.16)
v ≈ 5.7 + 245.83
v ≈ 251.53 m/s
So the velocity at the end of the acceleration is approximately 251.53 m/s.
To find the velocity after accelerating for a specific distance, you can use the equation:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled. Substituting the known values, you can solve for v:
v^2 = (5.7)^2 + 2(0.79)s
For 132 m:
v^2 = (5.7)^2 + 2(0.79)(132)
v^2 = 32.49 + 208.8
v^2 ≈ 241.29
v ≈ sqrt(241.29)
v ≈ 15.53 m/s
So the velocity after accelerating for 132 m is approximately 15.53 m/s.
For 65 m:
v^2 = (5.7)^2 + 2(0.79)(65)
v^2 = 32.49 + 102.7
v^2 ≈ 135.19
v ≈ sqrt(135.19)
v ≈ 11.62 m/s
So the velocity after accelerating for 65 m is approximately 11.62 m/s.